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In a process a system does 140j of work ...

In a process a system does 140j of work on the surroundings and only 40J of heat is added to the system. Hence change in internal energy is

A

180J

B

`-180J`

C

`-23.92` cal

D

`-100J`

Text Solution

Verified by Experts

The correct Answer is:
C, D
`Delta U= q + W_("on")`
`Delta U+ 40 - 140= -100J = -23.9` cal
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